9

I am trying to find the derivative of $f(x)= xe^x \csc x$, and I am not too sure how to even start.

Is it two terms or three? $xe^x$ and $\csc x$ or is it $x$, $e^x$ and $\csc x$? I can't get a proper answer either way.

With two terms I get $xe^x(-\csc x\cot x) + \csc x(e^x)$.

hardmath
  • 37,015
  • 1
    If you break it down as "two terms" (more precisely, two factors) --$x e^x$ and $\csc x$-- then when you get around to taking the derivative of the $x e^x$ factor, you need to recognize that it has two factors. The derivative of $x e^x$ is not $e^x$. – Blue Sep 22 '11 at 15:04
  • 1
    There's a neat shortcut to deriving the differentiation rule for products: note that the derivative of $\ln,f(x)$ is $\frac{f^\prime (x)}{f(x)}$, and consider what happens if $f(x)$ is a product of functions, using the fact that $\ln(ab)=\ln,a+\ln,b$... see this as well. – J. M. ain't a mathematician Sep 22 '11 at 15:04
  • What is ln? Is that short for natural log? Which is something raised to the 10th power? –  Sep 22 '11 at 15:07
  • 3
    Once you become fluent with the (two-factor) product rule, it would probably help to realize (and justify to yourself) that there's a many-factor product rule: for instance, with four factors, $(fghk)^\prime = f^\prime g h k + f g^\prime h k + f g h^\prime k + f g h k^\prime$. That is, you take the derivative of each factor in turn. – Blue Sep 22 '11 at 15:08
  • That makes sense to me. –  Sep 22 '11 at 15:10
  • @Jordan: I changed the title of the question - seems to fit more the question itself. – SBF Sep 22 '11 at 15:58

4 Answers4

17

To find the derivative of $(abc)'$ you use repeated application of the product rule: $$ (abc)' = (ab)'c+abc' = (ab'+a'b)c+abc' = a'bc+ab'c+abc'. $$ In your case $a(x) = x$, $b(x) = \mathrm e^x$ and $c(x) = \operatorname{csc}(x)$, so $$ a' = 1, b' = \mathrm e^x \text{ and }c' = -\cot x\csc x. $$

To make it more clear: in $x \mathrm e^x\csc x$ you have three function rather than two, but $x\mathrm e^x$ is also a product of two functions, so $$ (x\mathrm e^x\csc x)' = (x\mathrm e^x)'\csc x+x\mathrm e^x(\csc x)'. $$ We can calculate the latter term, but what about $(x\mathrm e^x)'$? You again apply the product rule: $$ (x\mathrm e^x)' = x'\mathrm e^x+x(\mathrm e^x)'. $$

SBF
  • 36,041
  • I don't know what that is, we are just suppose to be using the product rule. –  Sep 22 '11 at 15:02
  • Is it clear now? – SBF Sep 22 '11 at 15:10
  • Oh I didn't realize it was the chain rule, I am trying to use only the product rule for now incase it is on the test I have on monday. –  Sep 22 '11 at 15:12
  • 2
    @Gortaur, You're writing "chain rule" where you mean "product rule". And I think you mean "$x e^x$ is a product", not "... function". – Blue Sep 22 '11 at 15:15
  • 2
    @Jordan: I'm sorry - I studied this course in another language, so I cannot remember what is the chain rule. Clearly, in my answer I applied the only product rule, you can be sure that it's enough. Chain rule was just the misuse of the terminology. – SBF Sep 22 '11 at 15:16
  • @DayLateDon: you're certainly right. – SBF Sep 22 '11 at 15:18
  • 1
    @Jordan, You might want to bring up the many-factor product rule in your class. As Gortaur shows, it's the result of nothing more than applying the "regular" product rule over and over. Recognizing the pattern can greatly simplify your work. (Your classmates may hail you as a hero!) But before using the extended rule on a test, you'll want to clear it with your teacher, who (sadly) may be unaware of it, or who may be specifically testing you on writing out computations using the regular product rule. – Blue Sep 22 '11 at 15:25
  • 1
    Just to make sure I am getting this, is it then $(e^x + xe^x) (cscx) + -cscxcotx(xe^x)$? –  Sep 22 '11 at 15:26
  • @Jordan: yes... – SBF Sep 22 '11 at 15:34
5

As J.M. said, one easy way yo handle arbitrary products and quotients is to use the logarithmic derivative: $(\ln f(x))' =\dfrac{f'(x)}{f(x)} $.

Suppose $f(x)$ contains a mixture of products and quotients: $f(x) =\dfrac{a_1(x)a_2(x) ... a_m(x)}{b_1(x)b_2(x)...b_n(x)} =\dfrac{\prod_{i=1}^m a_i(x)}{\prod_{j=1}^n b_j(x)} $.

Taking the log, $\ln f(x) =\sum_{i=1}^m \ln a_i(x)-\sum_{j=1}^n \ln b_j(x) $.

Differentiating both sides,

$\begin{array}\\ \dfrac{f'(x)}{f(x)} &=(\ln f(x))'\\ &=(\sum_{i=1}^m \ln a_i(x)-\sum_{j=1}^n \ln b_j(x))'\\ &=\sum_{i=1}^m (\ln a_i(x))'-\sum_{j=1}^n (\ln b_j(x))'\\ &=\sum_{i=1}^m \dfrac{a_i'(x)}{a_i(x)}-\sum_{j=1}^n \dfrac{b_j'(x)}{b_j(x)}\\ \end{array} $.

Therefore

$\begin{array}\\ f'(x) &=f(x)\left(\sum_{i=1}^m \dfrac{a_i'(x)}{a_i(x)}-\sum_{j=1}^n \dfrac{b_j'(x)}{b_j(x)}\right)\\ &=\dfrac{\prod_{i=1}^m a_i(x)}{\prod_{j=1}^n b_j(x)}\left(\sum_{i=1}^m \dfrac{a_i'(x)}{a_i(x)}-\sum_{j=1}^n \dfrac{b_j'(x)}{b_j(x)}\right)\\ \end{array} $.

Here are the two most common cases:

$f(x) = a_1(x) a_2(x)$ - the product with $m=2$ and $n=0$.

$f'(x) =a_1(x) a_2(x)\left(\dfrac{a_1'(x)}{a_1(x)}+\dfrac{a_2'(x)}{a_2(x)}\right) =a_1'(x) a_2(x)+a_1(x) a_2'(x) $.

$f(x) = \dfrac{a(x)}{b(x)}$ - the quotient with $m=n=1$.

$f'(x) = \dfrac{a(x)}{b(x)}\left(\dfrac{a'(x)}{a(x)}-\dfrac{b'(x)}{b(x)}\right) = \dfrac{a'(x)}{b(x)}-\dfrac{a(x)b'(x)}{b^2(x)} = \dfrac{a'(x)b(x)-a(x)b'(x)}{b^2(x)} $.

Finally, your case - the product of three functions:

$\begin{array}\\ f(x) &=a(x)b(x)c(x)\left(\dfrac{a'(x)}{a(x)}+\dfrac{b'(x)}{b(x)}+\dfrac{c'(x)}{c(x)}\right)\\ &=a'(x)b(x)c(x)+a(x)b'(x)c(x)+a(x)b(x)c'(x)\\ \end{array} $

marty cohen
  • 107,799
1

$$ Let\quad y = \prod_1^n f_i(x)\quad and \quad thus\quad\ln(y) = \sum_1^n \ln(f_i(x))$$ recalling that$$ y'=y\space(\ln(y))'$$ then we obtain$$y'=\prod_1^n f_i(x)\space[\sum_1^n \ln(f_j(x))]'=\prod_1^n f_i(x)\space[\sum_1^n \dfrac {f'_j(x)}{f_j(x)} ]=\sum_1^n f'_i(x)[\prod_{j\neq i} f_j(x)] $$ and setting n=3 $$ (f_1 f_2 f_3)'=f'_1 f_2 f_3 +f_1 f'_2 f_3 + f_1 f_2 f'_3$$

Saul Mendoza
  • 231
0

From first principles, the product rule operation is invariant under the product of any number of functions. This can be proved through induction but it's interesting to see it using the definition of the derivative.

$$\lim_{dx \to 0}\frac{f_1(x+dx)...f_n(x+dx) - f_1(x)...f_n(x)}{dx}$$

is

$$\lim_{dx \to 0}\frac{(f_1(x)'dx+f_1(x))...(f_n'(x)dx+f_n(x)) - f_1(x)...f_n(x)}{dx}$$

Note that the only term in the expansion that doesn't end up with a $dx$ multiplied by it is the derivative of some $f_q$ multiplied by the product of the remaining functions. The product $f_1(x)f_2(x)...f_n(x)$ disappears in the numerator. It follows that the derivative is $$ {\sum_{p=1}^n\prod_{i=1}^n f_i(x)}\frac{f'_p(x)}{f_p(x)}$$.

for any product of functions $$\prod_{i=1}^n f_i(x)$$

jg mr chapb
  • 1,502
  • 1
  • 13
  • 44